Weakly atomic lattice

Question

The definition is that

The lattice $\mathcal{L}$ is said to be weakly atomic if whenever $a > b$ in $\mathcal{L}$, there exist elements $u,v \in L$ such that $a \geq u \succ v \geq b.$ ($\succ$ means cover).

I think all finite lattices are weakly atomic, is there a lattice that is not weakly atomic? Thanks in advance.

Answer 1

Indeed all finite lattices are weakly atomic.
For example $(\mathbb Q, \leq)$, where $\leq$ is the usual order, is not weakly atomic: whenever $u < v$ in $\mathbb Q$, there exists $w$ such that $u < w < v$ (just take $w = \frac{u+v}{2}$).